Question

I need help understanding integration by substitution: 63/(9x+2)^8

Answer 1

if you let
u = 9x+2
what is du
?

Answer 2

This is the as saying f(x) =9x+2
What is f'(x)
Just a difference of differentials but you don't need to worry about thst

Answer 3

du=9 right?

Answer 4

Yes just one thing you missed
Du=9dx

Answer 5

no du/dx =9

but if you "take the derivative"
u= 9x
you differentiate the variables, in this case the u and the x like this
du = 9 dx

Answer 6

and once you have
du = 9 dx
then you also have
\[ dx = \frac{1}{9} du \]

Answer 7

now do your variable substitution
replace 9x+2 with u
replace dx with 1/9 du

Answer 8

and if we have limits, we would also change the limits.

Answer 9

and because it is easy to integrate a power, I would use the -8 exponent and get rid of the fraction

Answer 10

So would it be: \[63u ^{-8}*1/9 du\] ??

Answer 11

yes and the problem is
\[ 7 \int u^{-8} du \]
which you can do , right?

Answer 12

Yes

Answer 13

once you integrate, replace u with 9x+2 to put the answer in terms of x

Answer 14

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{63}{(9x+2)^8}~dx}\)

\(\large\color{slate}{\displaystyle u=9x+2}\)

\(\large\color{slate}{\displaystyle du=(9x+2)'~\cdot dx~~~\rightarrow~~~du=9~dx}\)

(you already have a 9 and dx to replace that by du, just need to re-write it.

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{7\cdot 9}{(9x+2)^8}~dx}\)

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{7}{(9x+2)^8}~(9\cdot dx)}\)

substitution::

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{7}{(u)^8}~(du)}\)

and then all you have left::

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}7~u^{-8}~du}\)

Answer 15

(APPLY THE POWER RULE, AND don't forget to substitute the x back for u.)

Answer 16

igtg....

Answer 17

Ok, I got it! Thank you all!